Complete combustion of 1.11 0 g of a gaseous hydrocarbon yields 3.613 g of carbon dioxide and one 1.109 g of water. 80.288 g sample of the hydrocarbon occupies a volume of 131 milliliters at 24.8°c and 753 mm hg. what is the molecular formula of the hydrocarbon?

 C4H6

Explanation:

we know that

n = P×V / RT

= (753 ) x (0.131 ) / ((62.36367) x (24.8 + 273.15) ) = 0.005309

molecular weight = (0.288) / (0.005309) = 54.25 g/mol

molecular weight  of CO_2 in terms of C

(3.613 ) / (44.00964 ) x (12.01078) = 0.986033g C

Molecular weight of H_2O in terms of H

(1.109 ) / (18.01532 ) x (2  / 1) x (1.007947) = 0.124096g H

Number of atoms of C

(0.986033) / (1.110) x (54.25) / (12.01078) = 4.012

(0.124096) / (1.110) x (54.25 ) / (1.007947) = 6.017

Round to the nearest whole numbers to find the molecular formula:  C4H6

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